# Finite index subgroup of $\mathbb{Z}^2$ that is invariant under a non-singular matrix

Let $$M$$ be a matrix in $$\operatorname{GL}(2, \mathbb{Z})$$ that has at least one eigenvalue of absolute value strictly bigger than $$1$$. What are the finite index subgroups $$H$$ of $$\mathbb{Z}^2$$ such that for all $$v \in H$$, we have $$M(v) \in H$$?

Question: One example of such subgroup has the form $$H=\langle (a,0),(0,a)\rangle$$, for a non-zero integer $$a$$, this example works for any matrix in $$\operatorname{GL}(2, \mathbb{Z})$$. I was wondering if there are any other such subgroups.

Thought so far: I think the finite index subgroups of $$\mathbb{Z}^2$$ have the form $$H = \langle (a,b),(c,d)\rangle$$, where $$a,b,c,d \in \mathbb{Z}$$ such that $$ad-bc\neq 0$$. Hence, in order for $$H$$ to be a subgroup, we need $$M(a,b)= \alpha_1(a,b) + \alpha_2(c,d)$$ and $$M(c,d)= \beta_1(a,b) + \beta_2(c,d)$$ for some integers $$\alpha_i$$ and $$\beta_i$$, but I can't see any obvious solutions for these equations.

I was also thinking maybe we can use the eigenvectors, but unfortunately the eigenvectors might not have integers entries.

Any ideas for constructing such a subgroup would be really appreciated.

• Such a subgroup contains $n\mathbf{Z}^2$. So the question "reduces" to an understanding of determining, for all $n$, invariant subgroups of $(\mathbf{Z}/n\mathbf{Z})^2$ under the reduction of $M$ modulo $n$. So the answer should be mostly of arithmetic nature.
– YCor
Commented May 7, 2022 at 18:21
• The order of the quantifiers is not clear to me. Are you asking, for a given $M$, which subspaces $H$ are stable under it or, for a given $H$, whether it is stable under some $M$? Commented May 7, 2022 at 19:43
• @LSpice sorry, I meant to ask "for a given $M$, which subgroups $H$ are invariant under it" Commented May 7, 2022 at 19:55

If $$v$$ is not scalar, the ring $$\mathbb Z[M]$$ generated by $$M$$ has rank $$2$$ over $$\mathbb Z$$. It is therefore either an order in a number field, a finite-index subring of $$\mathbb Z^2$$, or $$\mathbb Z[\epsilon]/\epsilon^2$$.

$$\mathbb Z^2$$ is then a module over $$\mathbb Z[M]$$, and you are looking to describe the finite-index submodules of $$\mathbb Z[M]$$.

In the simplest case, $$\mathbb Z^2$$ is a (locally) free module over $$\mathbb Z[M]$$, and so you are describing the ideals of $$\mathbb Z[M]$$. In the ring of integers in a number field case, this has a particularly simple description, as products of prime ideals.

In the general order case, submodules are given by products of prime ideals away from those primes where the order $$\mathbb Z[M]$$ is not maximal or the module $$\mathbb Z^2$$ is not locally free, and there is additional complexity at those primes. Said another way, the simplest construction of such a subgroup is to take the product of $$\mathbb Z^2$$ with some ideal of the order $$\mathbb Z[M]$$, for example the intersection of $$\mathbb Z[M]$$ with an ideal of the ring of integers of the number field.

In the other cases there is a similarly concrete description.

Edit: Actually maybe I should say this a different way. Your idea of using the eigenvectors is a good one, but it's better to use the eigenvectors modulo $$p$$. Consider the characteristic polynomial of the element $$M$$, and its discriminant. If this discriminant is nonzero then there are infinitely many primes $$p$$ modulo which the discriminant is a nonzero square (quadratic reciprocity + Dirichlet's theorem). When the entries of $$M$$ are taken mod $$p$$, allowing us to view $$M$$ as a matrix over $$\mathbb F_p$$, it has two distinct eigenvalues and therefore two eigenvectors. The subgroup of vectors congruent mod $$p$$ to a multiple of one of these two eigenvectors is an index $$p$$ subgroup invariant under $$M$$.

If the discriminant is $$0$$ and $$M$$ is non-scalar, so $$M$$ has a single $$2 \times 2$$ Jordan block, then modulo all but finitely many primes $$p$$, $$M$$ will still have a single $$2\times 2$$ Jordan block, and you can take elements congruent mod $$p$$ to multiples of the unique eigenvector. Modulo the other primes, where $$M$$ becomes scalar, you can take multiples of any vector.

To illustrate how such a question is of arithmetic nature (and what a complete answer should look like), here is a partial answer in a specific case.

Namely: I specify to $$M=\begin{pmatrix}2 & 1 \\ 1 & 1\end{pmatrix}$$ and I address the question: what are invariant finite index subgroups of prime index.

The characteristic polynomial of this matrix is $$X^2-3X+1$$, which has discriminant 5. By the quadratic reciprocity formula, for an odd prime $$p\neq 5$$, 5 is a square mod $$p$$ iff $$p=\pm 1$$ mod $$5$$ (i.e. the decimal expansion of of $$p$$ terminates with $$1$$ or $$9$$). And this polynomial is irreducible mod $$2$$, and has a double root mod $$5$$.

Hence, for a prime $$p$$:

• for $$p=\pm 2$$ mod $$5$$, there is no $$M$$-invariant subgroup of index $$p$$ in $$\mathbf{Z}^2$$;
• for $$p=\pm 1$$ mod $$5$$, there are exactly two $$M$$-invariant subgroups of index $$p$$ in $$\mathbf{Z}^2$$;
• there is a single $$M$$-invariant subgroup of index $$5$$ in $$\mathbf{Z}^2$$.

Remarks:

For a general matrix $$M$$, the characteristic polynomial is $$X^2+nX\pm 1$$ for some $$n$$ and there should be a similar discussion.

For a general index $$q$$, one should boil down to when $$q$$ is a power of a prime $$p$$. Then in turn one should boil down to when the quotient is cyclic (the quotient being isomorphic to $$C_{p^a}\times C_{p^b}$$ for some $$a\le b$$, the subgroup is contained in $$p^a\mathbf{Z}^2$$ and we can then "replace" $$\mathbf{Z}^2$$ with $$p^a\mathbf{Z}^2$$ to assume $$a=1$$).

For A fixed $$M$$, such finite index subgroups may be categorized as follows. Let $$W\in GL(2,\mathbb{Z})$$ and $$b$$ a positive integer dividing the lower left element of the matrix $$W^{-1} M W$$. Then the columns of the matrix $$W\begin{bmatrix} 1 & 0\\0 & b\end{bmatrix}$$ generate a subgroup $$H$$ of $$\mathbb{Z}^2$$ (of finite index $$b$$) for which $$Mv\in H$$ for any $$v\in H$$. Conversely, all such subgroups for this $$M$$ arise this way, up to scaling the subgroup by an integer $$a$$ as already noted in the question.

To see this, first note that a finite index subgroup of $$\mathbb{Z}^2$$ may be generated by the columns of a non-singular integer matrix $$A$$, known as a basis. Any other basis equals $$AU$$, where $$U$$ is unimodular, i.e. an element of $$GL(2,\mathbb{Z})$$.

For unimodular $$M$$, you want to know for which such $$A$$ do we have $$MA = AV$$, for some integer matrix $$V$$, as such $$V$$ just makes integer linear combinations of the columns of the basis $$A$$, which give elements of the finite index subgroup $$H$$. But determinants then imply $$V$$ is also unimodular, as it has the same determinant as $$M$$. So $$M$$ takes any basis of $$H$$ to another basis of $$H$$.

Integer matrices have a Smith Normal Form: there exist unimodular $$W,V$$ so that $$W\begin{bmatrix} a & 0\\0 & ab\end{bmatrix}V = A$$, where $$a,b$$ are integers (here positive, as the determinant is non-zero). As any choice of basis will do, we choose the basis $$W\begin{bmatrix}a & 0\\0 & ab\end{bmatrix}$$.

So an $$H$$ may be specified by a unimodular $$W$$ and integers $$a$$ and $$b$$. As $$a$$ simply scales everything we henceforth assume WLOG that $$a=1$$.

We now characterize such pairs $$W, b$$. We have from before the equation $$M (W\begin{bmatrix} 1 & 0\\0 & b\end{bmatrix}) = (W\begin{bmatrix} 1 & 0\\0 & b\end{bmatrix}) V$$ so $$(W^{-1} M W) \begin{bmatrix} 1 & 0\\0 & b\end{bmatrix} = \begin{bmatrix} 1 & 0\\0 & b\end{bmatrix} V$$

As $$W$$ is unimodular, it has integer inverse, so all matrices above are integer. The matrix on the left has second column divisible by $$b$$, while the matrix on the right has second row divisible by $$b$$. As they equate, both sides must equal a matrix of the form $$\begin{bmatrix} x & by\\bz & bw\end{bmatrix}$$ where $$w,x,y$$ and $$z$$ are integers. But then $$V$$ equals $$\begin{bmatrix} x & by\\z & w\end{bmatrix}$$, so $$V$$ unimodular implies $$xw - byz = \pm 1$$.

Similarly, we have $$W^{-1} M W = \begin{bmatrix} x & y\\bz & w\end{bmatrix}$$, since the diagonal matrix on the right simply scales the second column by $$b$$. So we started with a general subgroup $$H$$ invariant under $$M$$ - generated by the columns of a matrix $$A$$ - and showed that this $$H$$ has a basis of the form $$W\begin{bmatrix} 1 & 0\\0 & b\end{bmatrix}$$ where the index $$b$$ divides the bottom left entry of $$W^{-1} M W$$, as claimed.

For the other direction, let $$b$$ be a positive factor of the bottom left entry of $$W^{-1} M W$$, where $$W$$ is any unimodular matrix. This means

$$W^{-1} M W = \begin{bmatrix} x & y\\bz & w\end{bmatrix}$$

for some integers $$w, x, y$$ and $$z$$, where $$xw - bzy =\pm 1$$, as the matrix is unimodular. But then $$M W\begin{bmatrix} 1 & 0\\0 & b\end{bmatrix} = W \begin{bmatrix} x & y\\bz & w\end{bmatrix}\begin{bmatrix} 1 & 0\\0 & b\end{bmatrix}$$ and $$M \left(W\begin{bmatrix} 1 & 0\\0 & b\end{bmatrix}\right)= W\begin{bmatrix} x & by\\bz & bw\end{bmatrix}=W \begin{bmatrix} 1 & 0\\0 & b\end{bmatrix} \begin{bmatrix} x & by\\z & w\end{bmatrix}= \left(W\begin{bmatrix} 1 & 0\\0 & b\end{bmatrix}\right) V$$

where the matrix $$V$$ is unimodular, as we have already noted $$xw - bzy =\pm 1$$. But this means the subgroup $$H$$ of $$\mathbb{Z}^2$$ generated by the columns of $$W\begin{bmatrix} 1 & 0\\0 & b\end{bmatrix}$$ is invariant under $$M$$, as claimed.